# The Chain Rule – An Integral Part of Calculus

## Chain Rule Calculus

Hey guys! Welcome to this video on how to differentiate using the chain rule.

At this point you should should know how to take the derivative of functions like:

F(x)=Sqrt(x)   , R(t)= cos(t)      , Z(p)= e^p     , Y(n)=n^24


These are somewhat straightforward. We can use use the power rule, the quotient rule, or the product rule.

But, what about a function like this:

Y(t)=sqrt(9t+5)

Using the definition of a derivative to take the derivative we get Y’(t)=9/2(sqrt(9t+5))^-1/2.

If we use the power rule to try and take the derivative we get Y’(t)=1/2(sqrt(9t+5))^-1/2.

These two functions may look similar, but as you can see they’re not. We have different numbers here in the numerator. We could just use the definition of a derivative to take the derivative of a function every time, but if we take a close look at this function (Y(t)=sqrt(9t+5)), we could say that f(t) is equal to sqrt(g(t)) and our g(t) is equal to everything on the inside here (9t+5). Then what we could do is write it as a composition function.

Y(t)=(f०g)(t)=(f(g(t)))=sqrt(9t+5)

Well, this sets us up then to be able to use the chain rule.

There are two types of notations in which you can choose to use the chain rule, and I will list both.

Let’s say you have two functions f(x) and g(x) and they are both differentiable.

You could define Z(x)=(f०g)(x) to make the derivative of Z(x):

Z’(x)=f’(g(x))*g’(x)

The other type of notation is called Leibniz’s notation.

If you have the function y=f(u) and u=g(x) then the derivative of y is
dy/dx=(dy/du)*(du/dx).

In this video I will use the first type of notation, or the prime notation.

Now, let’s look at how to apply the chain rule.

Use the Chain rule to differentiate Y(t)=sqrt(9t+5).

We’ve already looked at this problem, and figured out which two functions are needed for the composition.

So, we can go ahead and take the derivatives of these two functions.

f(t)=sqrt(g(t))    f’(t)= 1/(2(sqrt(t))^-1/2
g(t)=(9t+5)        g’(t)= 9
Then, using the chain rule we get:
Y’(t)=f’(g(t))g’(t)
=(1/(2(sqrt(9t+5)^-1/2)*9
=9/2(sqrt(9t+5))


One thing you’ll notice, and I think is super helpful, is that I will always write my two functions that my original function is made up of, and both of their derivatives. This helps me to keep track of my work.

Alright, let’s take a look at another example.

Differentiate the following function.

B(h)= 7(5h^3 + cos(h))6

Now, let’s stop and take a look at our function. What two functions is this function made up of?

Well, we have 7 times everything on the inside here to the 6th power. Then, we have the inside function here which is 5h^3 + cos(h). So, we can break this function up into two separate functions.

I’m going to set f(x) equal to 7h^6 because it is my outside function; and I’m going to set g(x) equal to 5h^3 + cos(h), because it is my inside function. Now, we take the derivative of each of our two functions.

      f’(x)=42h^5 ,  g’(x)=15h^2 -sin(h)


So, we are now ready to use the chain rule. Another helpful way to think through how to use and apply the chain rule is this. The derivative of the outside function, times the inside function left alone, times the derivative of the inside function. Now, let’s walk through that.

So, B’(h) is equal to the derivative of the outside function. So, that is our f’(x). Then, we multiply that times the inside function left alone. So, that is our g(x). Finally, we multiply all of this times the derivative of the inside function. So, that is g’(x).

B’(h)=42(5h^3 + cos(h))^5 * 15h^2 -sin(h)

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by Mometrix Test Preparation | Last Updated: April 15, 2020